How many grams of MgCl2 will be obtained when 435 mL of 0.300 M HCl react with an excess of Mg (OH) 2?

The grams of Mg (OH) 2 produced is calculated as below

calculate the moles of HCl produced = molarity xvolume/1000

= 0.3 x 435/1000 = 0.1305 moles

write the equation for reaction

Mg (OH) 2 + 2HCl = MgCl2 + 2H2O

by use of mole ratio between HCl : MgCl2 which is 2 : 1 the moles of HCl = 0.1305 x1/2 = 0.0653 moles of MgCl2

mass of MgCl2 = moles x molar mass

= 0.0653mol x95 g/mol = 6.204 grams of MgCl2